# Chapter 28 Problem 49 † Given C = 0.15 µF L = 20 mH R = 1.6 Ω

advertisement ```Chapter 28
Problem 49
†
Given
C = 0.15 &micro;F
L = 20 mH
R = 1.6 Ω
Solution
Find the number of cycles it takes to have the peak voltage drop to half its original value.
The formula describing charge on a capacitor in a damped oscillator circuit is
q = qo e−Rt/2L cos ωt
Charge on a capacitor is related to the voltage on the capacitor by
V = CQ = Cqo e−Rt/2L cos ωt = Vo e−Rt/2L cos ωt
Ignoring the cosine function, which does not cause the voltage in the circuit to decay, we can determine
the time for the voltage to decay to half its value.
V
= e−Rt/2L
Vo
−2L
V
t=
ln
R
Vo
Substituting in the appropriate values gives us a time of
1
−2(0.020 H)
ln
t=
= 0.0173 s
1.6 Ω
2
The angular frequency of the circuit is
ω=√
1
1
=p
= 1.83 &times; 104 rad/s
LC
(0.020 H)(1.5 &times; 10−7 F )
This gives a time period for each oscillation of
T =
2π
2π
=
= 3.44 &times; 10−4 s
ω
(1.83 &times; 104 rad/s)
The number of cycles before reaching 1/2 voltage is then
1cycle
t = 0.0173 s
= 50.3 cycles
3.44 &times; 10−4 s
†
Problem from Essential University Physics, Wolfson
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